so i did a wolfram alpha check and the solution looks nasty... so there probably is a closed form method but its likely pretty involved & tedious, after looking at it closer, I can't see any easy way to simplify...

getting late, so i've probably made a mistake, but see what you think of this logic...
assume x>0 and hopefully we don't divide by zero anywhere
[tex] \sqrt{x}(\sqrt{A-x^2}+\sqrt{B-x^2})=C[/tex]
multiply by the rational to get
[tex] \sqrt{x}(A-B)=C(\sqrt{A-x^2}-\sqrt{B-x^2})[/tex]

re-arrange both to get
[tex] \sqrt{A-x^2}+\sqrt{B-x^2}=\frac{C}{\sqrt{x}}[/tex]
[tex] \sqrt{A-x^2}-\sqrt{B-x^2}=\frac{\sqrt{x}(A-B)}{C}[/tex]

adding and subtracting we get to
[tex] 2\sqrt{A-x^2}=\frac{C}{\sqrt{x}}+\frac{\sqrt{x}(A-B)}{C}[/tex]
[tex] 2\sqrt{B-x^2}= \frac{C}{\sqrt{x}}-\frac{\sqrt{x}(A-B)}{C}[/tex]

In the original equation, solutions exist only for positive C, and both sides of the equation are trivially positive where solutions exist (in the real numbers).

Given this, square both sides of the equation; since if f(x) = g(x) and both f(x) and g(x) are positive, then [f(x)]^2 = [g(x)]^2 iff f(x) = g(x).

Rearrange terms so that only the radical term (only one remains after squaring) is on the right, and everything else is on the left of the equation. Clearly the right-hand side is positive, so solutions exist only when the left-hand side is positive... we must check our answers for x to ensure that this condition is satisfied. For now, square again.

Combine all like terms, and you arrive (unless I goofed) at an equation like this:

(4C^2)x^3 + (A^2 + B^2 - 2AB)x^2 - 2C^2(A + B)x + C^4 = 0.

This is a cubic equation in standard form, and since there is a method to solve cubics, this can be solved. Simply check the answers against the conditions we have identified and that should contain all answers...

If you want to solve it yourself without the formula, I suggest the following:
- do a linear transformation of x to remove the coefficient of x^2
- substitute x = y+a/y for a suitable constant a to transform it into a second-degree polynomial equation in y^2

Note that we may have generated more solutions that there are by squaring, so make sure that you are finding the correct ones.